Question: Solve for $x$ : $6x^2 - 108x + 480 = 0$
Answer: Dividing both sides by $6$ gives: $ x^2 {-18}x + {80} = 0 $ The coefficient on the $x$ term is $-18$ and the constant term is $80$ , so we need to find two numbers that add up to $-18$ and multiply to $80$ The two numbers $-10$ and $-8$ satisfy both conditions: $ {-10} + {-8} = {-18} $ $ {-10} \times {-8} = {80} $ $(x {-10}) (x {-8}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x -10) (x -8) = 0$ $x - 10 = 0$ or $x - 8 = 0$ Thus, $x = 10$ and $x = 8$ are the solutions.